3. homework n1

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Example 5 : RETAINING WALL BASE DESIGN

Using the Eurocode design philosophy and the provisional buried-footinginteraction diagrams deduce: (i) an acceptable breadth for the base with aburial depth D = 0; (ii) reduce this arbitrarily by 2m and use the above

interaction diagrams to arrive at a plausible burial depth D for this case.Although the (V,H) forces will generate moments M about an axis through thecentreline of the base you will find that they are (or can be) quite small andtherefore a 2D (V,H) interaction diagram is adequate for your purpose.

[You might also like to answer the same questions using any traditionalapproach, the NGI charts or information in Powrie or Bowles etc.]

Assume that the soil involved is granular φ’ = 30º (the Eurocode (4/5)tanφ’material strength reduction factor leads to φ’ = 25º for use in ‘failure ‘calculations). The wall-fill is drained, γ =18 kN/m3 and the base is well aboveGWL, so no water pressure effects need to be considered in this case and h =

8m.

Use the bold outline dimensions in your calculations and ignore the weights ofthe concrete wall stem and the base. The forces transmitted to the wall-baseare: V = γ a h and H = ka γ h2 /2.

The simplest procedure for you to follow is probably:a. Select a trial value for B and, 3 or 4 possible pairs of (a,b) values.b. Calculate the (V, H) loads. H will be constant throughout.c. From these deduce the equivalent resultant force system referred to themid-point of the underside of the base (Vo, Ho, Mo/B), say.d. Calculate Nq and Nγ, hence Vmax (for D= 0). .e. Draw the D = 0 failure envelope (interaction diagram) and add the calculatedvalues of (Vo, Ho, Mo/B)/ Vmax. At failure they will lie on (or close to) the

envelope. The effect of (Mo/B)/Vmax is assumed to be close to zero.f. REPEAT this process with different values of B until success is achieved!

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For the second part of the exercise reduce the successful B (for D= 0) value to(B - 1.5)m. Repeat the above process (B is now FIXED) for a range ofconsistent (a, b) and (D/B) = (0.25, 0.5,1.0) [note that, in these diagrams, Vmaxis assumed to be a linear function of D/B ].

Also note that the moment axis and therefore the force origin,should now be located NOT at footing base level but at adepth of 2D/3 (WHY?).Draw the set of ‘shallow-buried-footing’ parabole and plot your calculated(V/Vmax, H/Vmax) results to estimate the minimum acceptable value of D = d.B.

IF you are’ enthused’ by all this, you could do a second similar set ofcalculations to determine a B (and a D/B) value for a ‘hazard’ in which the walldrainage becomes blocked, resulting in the water level being raised to 5mabove the wall base. A reasonable flow net under the wall (you might checkthis too?) provides an uplift pore-wter pressure distribution that varies,

approximately linearly, across the base from 5m of water to zero adjacent tothe lateral-resistance “downstand”. There will also be an additional horizontalwater pressure force acting on the wall: so the force system becomes a bitmore complicated!

IF YOU DO EXPLORE THIS PROBLEM YOU WILL FIND THAT SUCH A DRAINBLOCKAGE WOULD BE DISASTROUS --- THIS IS THE MOST COMMON CAUSEOF WALL COLLAPSE!!

General Remarks

1. If the equation to the full (V, H, M/B) envelope is known (as it is for the D = 0

case) then, for the wall footing , it can be expressed in terms of (a, b) as theonly unknowns. Computer packages can find roots of this equation (cigar==0,say), within specified ranges of (a and b), that lie precisely on the surface ofthe ‘cigar’.

2. Further than this, a more sophisticated analytical package allows you topose the problem in a form such as: Minmize[a+b, cigar==0, {a,b}] whichprovides the ‘exact’ minimum values of a and b that satisfy cigar==0.

3. This can be done for the buried footing cases if, for example, we make theFURTHER assumption that the cross-sections of them are all similarly rotatedand have the same axis ratios. The problem can now be posed as one in

‘linear algebra’ to find a minimum D = d.B value in a form such as:Minimize[ (a+b) d, {buriedcigar == 0, list ranges for a,b,d}, { a,b,d}].The ‘cost function’ (a+b) d can also be modified to take account of the fact thatburied ‘downstand’ construction is much more expensive than a ground-bearing slab e.g. Minimize[(a+b) + 3 (a+b) d …

4. A piled-base may provide a cheaper solution – especially if there arerestrictions on the breadth of footing that can be accommodated efficiently.

5 IT IS VITAL TO REMEMBER that, in a design office, ALL yourcalculations and conclusions WILL be checked by someone

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else. Therefore they MUST ALWAYS be presented CLEARLYand CONCISELY!!

6. COMPLETING ALL OF THE ABOVE IS QUITE A

DEMANDING TASK AND THEREFORE A MINIMALLYADEQUATE SUBMISSION DOES NOT HAVE TO INCLUDE AFULL ANALYSIS OF EITHER: (i) THE INUNDATED WALL OR(ii) A TRADITIONAL DESIGN OF THE BASE.

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PROCEDURES FOR CHECKING (V,H,M/B)/Vmax ON ARETAINING WALL BASE OVER A RANGE OF PARAMETERS

COURSEWORK EXAMPLE – supplementary notes:

I though that our discussion of this problem did not end with an adequateexplanation of how to proceed. These notes are intended to rectify this.

H = (γ ka h2)/2

h

V = γ a hh/3

B D = d.B

a b

PROBLEM INPUT: wall height (h = 8m for example), soil unit weight, soilfriction angle,

WALL & BASE WEIGHTS ARE NEGLECTED – including them, as a refinement,will increase V and also M (slightly).

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B = (a+b) is to be found for, if possible, a failure-state surface footing for thewall using a reduced value of φ = 25deg, otherwise a ‘downstand’ will beneeded.

1. One approach is simply to calculate the loads, for a trial value of B = 9m(say) and a range of (a, b) values to find a set of results that best fit the failurecriteria; (V/Vmax <= 1, H/Vmax <= .125, M/BVmax <= ,09).

The following output from this process suggests that any of them couldprovide a solution [PLOT THEM] -- since M/BVmax is very small

a b Vert êVmax HorêVmax HMomêBLêVmax

7.5 1.5 0.328814 0.106762 0.0200484

7. 2. 0.306893 0.106762 0.0133504

6.5 2.5 0.284972 0.106762 0.00787012

6. 3. 0.263051 0.106762 0.00360772

5.5 3.5 0.24113 0.106762 0.000563146

These are the corresponding results using φ = 30degrees ; obviously they willalso be OK.

a b Vert êVmax HorêVmax HMomêBLêVmax

7.5 1.5 0.147462 0.0393232 0.00518847

7. 2. 0.137631 0.0393232 0.00218462

6.5 2.5 0.1278 0.0393232 −0.000273077

6. 3. 0.117969 0.0393232 −0.00218462

5.5 3.5 0.108139 0.0393232 −0.00355001

The dimensionless (H/Vmax, V/Vmax) parabolic failure locus is plotted belowtogether with the (a, b) = (6.5 , 2.5) [i.e. B = 9m] load points for φ = 25 (in blue)and φ = 30 (in red).NB The locus clearly needs to be rather larger, try the (a, b) = (7.5, 1.5) solutionor perhaps B should be increased to 9.5m, say : SEE BELOW

0.2 0.4 0.6 0.8 1

0.02

0.04

0.06

0.08

0.1

0.12

2. IF THE FULL 3D FAILURE LOCUS IS KNOWN THEN an 'accurate' solution

can be obtained directly by calculating failure points that lie exactly on thecigar .

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THIS RESULT IS SHOWN BELOW FOR THE D = 0 CASE .(B, a, b) are now direct OUTPUT not estimated input

The (known) equation to the 3D, surface footing, cigar is:

(11 M/(Vmax B))^2 + (8 H/Vmax)^2

+ 0.44*88*H* M/(Vmax^2*B) - (4(V/Vmax)*(1 - V/Vmax))^2 = 0 (1)

(You can check this by either, plotting it, checking the parabolas when H= 0 orM/B = 0, or by plotting the largest, inclined elliptic cross-section when (V/Vmax= 0.5)), A wall-loading case lies in the upper ‘small’ quadrant of the inclinedellipse, A non-inclined ellipse has no H*M term.

When (V, H , M/B), for specific values of h and soil properties, are expressed interms of (a, b) [ as also done in section 1] this equation can be solved to findpairs of (a, b) values that lie precisely on the inclined cigar. Typical output is

shown below from which (7.75, 1.5) and (7.25, 2.0) are possible ‘exact’solutions. [NOW cf. this approach with any more conventional solution methodbased various multiplying factors!]

b a B

1 8.30491 9.30491

1.5 7.7453 9.2453

2. 7.26626 9.26626

2.5 6.86459 9.36459

3. 6.53448 9.53448

3.5 6.26836 9.76836

4. 6.05796 10.058

3. The final stage in this progression would be to treat equation(1), when (V, H, M/B) are expressed in terms of (a, b) , as a constraint in a minimisationprocedure to minimise (a+b).i.e. find (a+b)min subject to ‘equation(1) as a function [a,b]’. Packages areavailable which do such operations, providing directly, as an optimumsolution: B = 9.24m; a = 7.62m; b = 1.62m

NB: IF THERE IS INSUFFICIENT SPACE in-situ FOR A FOOTING WITHa = 7.6 m THEN, SINCE WE KNOW THAT a = 4m IS TOO NARROW FORSURFACE FOOTING A SOLUTION WITH AN INCREASED b AND A

'DOWNSTAND' WILL BE REQUIRED (or a piled foundation etc !).

When a full 3D model, incorporating M/B loading, becomesavailable for buried footings a direct solution could be foundfor this case also, seeking (a, b, D) essentially as in 3 above.See also the development for buried footings in the attachedMathematica procedure.

RB 13/10/2012

3. homework n1

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